Read the pH: below 7.35 is acidosis, above 7.45 alkalosis.
Find the value that explains it: PaCO₂ moving opposite to the pH is respiratory, HCO₃⁻ moving with the pH is metabolic.
Look at the other value: normal means uncompensated, abnormal in the compensating direction means partial compensation, and a normal pH with both abnormal means full compensation.
In a metabolic acidosis, compare PaCO₂ with Winter’s formula, 1.5·HCO₃⁻ + 8 ± 2, and calculate the anion gap Na⁺ − (Cl⁻ + HCO₃⁻).
Check a high anion gap with the delta ratio and, when albumin is low, with the corrected gap.
ECG rhythm strip interpretation
Paper (25 mm/s, 10 mm/mV)
1 small box = 0.04 s, 1 large box = 0.20 s, 10 mm = 1 mV
Heart rate, regular rhythm
300 ÷ large boxes = 1500 ÷ small boxes = 60 ÷ RR (s)
Heart rate, irregular rhythm
QRS complexes in 6 s × 10
Normal PR interval
0.12–0.20 s
Normal QRS duration
< 0.12 s
QTc (Bazett)
QT/√RR
QTc (Fridericia)
QT/∛RR
QTc (Framingham)
QT + 154·(1 − RR)
Sinus rates
bradycardia < 60, normal 60–100, tachycardia > 100 /min
Escape rates
junctional 40–60, ventricular 20–40 /min
Fast rhythms
SVT 150–250 /min (narrow), VT > 100 /min (wide)
Method
Rate: for a regular rhythm divide 300 by the large boxes or 1500 by the small boxes between two R waves; for an irregular rhythm count the QRS complexes in 6 seconds and multiply by 10.
Rhythm: compare the R–R intervals. Equal is regular; a repeating pattern is regularly irregular; no pattern at all is irregularly irregular.
P waves: is there an upright P wave in lead II before every QRS, all alike? Look for fibrillatory or sawtooth waves, inverted P waves and P waves without a QRS.
PR interval: from the start of P to the start of the QRS, normally 0.12–0.20 s and constant. Watch whether it lengthens, stays fixed before a dropped beat, or varies at random.
QRS duration: below 0.12 s is narrow; 0.12 s or more is wide. Then name the rhythm, and for the QT use QTc = QT/√RR (Bazett) with times in seconds.
Evidence-based practice statistics
Sensitivity and specificity
Se = a/(a + c), Sp = d/(b + d)
Predictive values
PPV = a/(a + b), NPV = d/(c + d)
Accuracy
(a + d)/n
Likelihood ratios
LR+ = Se/(1 − Sp), LR− = (1 − Se)/Sp
PPV at prevalence p
Se·p / (Se·p + (1 − Sp)(1 − p))
Odds
odds = p/(1 − p); p = odds/(1 + odds); post-test odds = pre-test odds × LR
Risk measures
RR = EER/CER, ARR = CER − EER, RRR = ARR/CER
Number needed to treat
NNT = 1/ARR, rounded up; NNH = 1/(EER − CER)
Odds ratio
OR = ad/(bc)
Incidence rate
new cases / person-time
Method
Draw the 2 × 2 table and label a, b, c and d before you calculate.
Read sensitivity and specificity down the columns, predictive values along the rows.
For another prevalence, imagine 10 000 people and fill the table with Se, Sp and the prevalence.
For a likelihood ratio, convert the probability to odds, multiply, and convert back.
For a trial, find EER and CER first; then RR = EER/CER, ARR = CER − EER and NNT = 1/ARR rounded up.
2Practice problems
1.Human anatomyMedium
Which structure produces most of the cerebrospinal fluid?
APia mater
BArachnoid granulations
CPineal gland
DChoroid plexus
2.Human anatomyHard
Which coronary artery most often supplies the sinoatrial (SA) node?
ARight coronary artery
BLeft main coronary artery
CLeft anterior descending artery
DCircumflex artery
3.Human physiologyMedium
Cardiac output equals heart rate multiplied by:
AStroke volume
BBlood pressure
CEjection fraction
DPeripheral resistance
4.Human physiologyHard
According to the length–tension relationship, why does a skeletal muscle fibre produce less force when overstretched?
AActin and myosin overlap less, so fewer cross-bridges form
BCalcium is no longer released from the sarcoplasmic reticulum when stretched
CThe muscle runs out of ATP when stretched
DThe neuromuscular junction stops working
5.MicrobiologyMedium
In which phase of the bacterial growth curve do the cells divide at their fastest constant rate?
ADeath phase
BLog (exponential) phase
CLag (adaptation) phase
DStationary (plateau) phase
6.MicrobiologyHard
Why is Chlamydia trachomatis usually diagnosed with nucleic acid amplification tests instead of routine culture on agar?
AIt is too large to grow on agar plates
BIt dies immediately in the presence of oxygen
CIt is a virus and has no DNA
DIt is an obligate intracellular bacterium that cannot grow on ordinary media
7.PharmacologyMedium
Why are many drugs started at lower doses in older adults?
ALower doses are cheaper for older patients
BOlder adults absorb no drugs from the stomach
COlder adults always weigh more than younger adults, so drugs last longer
DKidney and liver clearance often decline, and sensitivity to drugs increases
8.PharmacologyHard
Why is strict aseptic technique and a limited hang time needed for propofol infusions?
APropofol breaks down into a toxic gas after a few hours
BLight exposure changes propofol into an opioid
CIts lipid emulsion lets bacteria grow if it is contaminated
DPropofol becomes more potent the longer it hangs
9.Arterial blood gas interpretationMedium
A practice gas shows a metabolic acidosis with HCO₃⁻ 12 mmol/L. Use Winter’s formula to find the expected PaCO₂ (the centre of the range) in mmHg.
HCO₃⁻ 12 mmol/L
PaCO₂ = mmHg
10.Arterial blood gas interpretationHard
At sea level, a practice gas taken on 24% oxygen shows PaO₂ 53 mmHg (7.1 kPa) and PaCO₂ 45 mmHg (6.0 kPa). Use PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 to find the A–a gradient.
Measure the QT interval and the R–R interval on this regular lead II strip, then calculate the rate-corrected QTc with Fridericia’s formula QTc = QT/∛RR (both in seconds). Give the answer in seconds to three decimals.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
QTc = s
12.ECG rhythm strip interpretationHard
Identify the rhythm on this lead II strip. Click your answer.
lead II, 25 mm/s, 10 mm/mV1 small box = 0.04 s, 1 large box = 0.20 sstrip length 6 s
1sinus bradycardia
2asystole
3idioventricular rhythm
4junctional rhythm
13.Evidence-based practice statisticsMedium
Practice exercise (made-up values, not clinical guidance). A case-control study compares 210 patients with a disease (cases) and 420 people without it (controls). The table shows how many in each group were exposed to a risk factor. Find the odds ratio OR for the exposure. Round to two decimal places.
a = 157b = 227c = 53d = 193
OR =
14.Evidence-based practice statisticsHard
Practice exercise (made-up values, not clinical guidance). A patient has a pre-test probability of 5% for a disease. A test with sensitivity 63% and specificity 93% comes back negative. Find the post-test probability p′ of the disease as a percentage. Round to one decimal place.
pre-test 5%Se = 63%Sp = 93%negative
p′ = %
3Solutions
1.Human anatomy
Answer: D · Choroid plexus
Why this is the answer
The choroid plexus in the ventricles secretes most CSF. CSF is reabsorbed into venous blood through the arachnoid granulations.
Answer
D · Choroid plexus
2.Human anatomy
Answer: A · Right coronary artery
Why this is the answer
The SA nodal artery arises from the right coronary artery in roughly 60% of people and from the circumflex artery in the rest.
Answer
A · Right coronary artery
3.Human physiology
Answer: A · Stroke volume
Why this is the answer
CO = HR × SV. At rest about 70 beats/min × 70 mL ≈ 5 L/min.
Answer
A · Stroke volume
4.Human physiology
Answer: A · Actin and myosin overlap less, so fewer cross-bridges form
Why this is the answer
Maximum force occurs at an optimal sarcomere length with best filament overlap. Overstretching reduces overlap; too short a length makes filaments collide.
Answer
A · Actin and myosin overlap less, so fewer cross-bridges form
5.Microbiology
Answer: B · Log (exponential) phase
Why this is the answer
After the lag phase of adjustment, cells enter the log phase and grow exponentially. Growth stops in the stationary phase as nutrients run out and waste builds up.
Answer
B · Log (exponential) phase
6.Microbiology
Answer: D · It is an obligate intracellular bacterium that cannot grow on ordinary media
Why this is the answer
Chlamydia replicates only inside host cells, so it cannot be grown on agar. NAATs such as PCR on urine or swabs are sensitive and are the standard diagnostic test.
Answer
D · It is an obligate intracellular bacterium that cannot grow on ordinary media
7.Pharmacology
Answer: D · Kidney and liver clearance often decline, and sensitivity to drugs increases
Why this is the answer
Ageing reduces kidney function, changes body composition and often slows liver metabolism. Many drugs also have stronger effects on the ageing brain, so doses are started low and increased slowly.
Answer
D · Kidney and liver clearance often decline, and sensitivity to drugs increases
8.Pharmacology
Answer: C · Its lipid emulsion lets bacteria grow if it is contaminated
Why this is the answer
Propofol is formulated in a soybean oil emulsion, which is a good growth medium for bacteria. Contaminated propofol has caused sepsis, so it is handled aseptically and tubing and containers are changed at set intervals.
Answer
C · Its lipid emulsion lets bacteria grow if it is contaminated
9.Arterial blood gas interpretation
Answer: PaCO₂ = 26 mmHg
Expected compensation: in metabolic acidosis Winter’s formula predicts PaCO₂ = 1.5·HCO₃⁻ + 8 ± 2 mmHg; a measured value above that range points to an additional respiratory acidosis, below it to an additional respiratory alkalosis. The anion gap AG = Na⁺ − (Cl⁻ + HCO₃⁻), normally about 8–12 mmol/L, separates high-gap from normal-gap metabolic acidosis; low albumin hides a high gap (add 2.5 per g/dL that albumin is below 4.0). The delta ratio (AG − 12)/(24 − HCO₃⁻) between about 0.8 and 2 fits a pure high-gap acidosis (some texts use 1 instead of 0.8), below that an added normal-gap acidosis, and above 2 an added metabolic alkalosis.
Winter’s formula
PaCO₂ = 1.5·12 + 8 = 26 mmHg
range: 24–28 mmHg
Answer
PaCO₂ = 26 mmHg
10.Arterial blood gas interpretation
Answer: A–a gradient = 61.87 mmHg
Derived values: calculated serum osmolality is 2·Na⁺ + glucose + urea with all values in mmol/L (2·Na⁺ + glucose/18 + BUN/2.8 with glucose and BUN in mg/dL); normal is about 275–295 mOsm/kg. The P/F ratio is PaO₂ divided by FiO₂ as a fraction. The alveolar oxygen pressure is PAO₂ = FiO₂·(760 − 47) − PaCO₂/0.8 mmHg at sea level, and the A–a gradient is PAO₂ − PaO₂.
QT and QTc: the QT interval runs from the start of the QRS complex to the end of the T wave. It shortens as the heart rate rises, so it is corrected for rate with the R–R interval in seconds: Bazett QTc = QT/√RR and Fridericia QTc = QT/∛RR. Bazett is the formula most textbooks use; it over-corrects at fast rates, where Fridericia is preferred. Limits differ between references: a QTc above about 0.45 s in men or 0.46 s in women is usually called prolonged, and above 0.50 s markedly prolonged.
Measure
QT = 12 × 0.04 = 0.48 s
RR = 25 × 0.04 = 1.00 s
Fridericia’s formula
QTc = 0.48/∛1.00 = 0.48/1 = 0.48 s
Answer
QTc = 0.48 s
Above the usual limits for both men (about 0.45 s) and women (about 0.46 s): prolonged. (480 ms)
12.ECG rhythm strip interpretation
Answer: idioventricular rhythm
More ventricular rhythms: torsades de pointes is a polymorphic ventricular tachycardia, usually 150–250/min, in which the wide complexes change in height and twist around the baseline, pointing up for a few beats and then down. Textbooks link it to a prolonged QT interval, and magnesium sulfate is the drug resuscitation guidelines name for it. An idioventricular rhythm is a ventricular escape rhythm: regular, wide QRS (0.12 s or more), no P waves, at 20–40/min (an accelerated idioventricular rhythm runs at 40–100/min). Monomorphic VT has identical wide complexes at more than 100/min. VF is chaotic, without any repeating complex.
Five steps
1. Rate: R–R = 54 small boxes, 1500 ÷ 54 ≈ 28 per minute.
2. Rhythm: regular, every R–R is 2.16 s.
3. P waves: none identifiable.
4. PR interval: not measurable.
5. QRS: 0.18 s, wide (0.12 s or more) and bizarre, with the T wave pointing the other way.
Interpretation
idioventricular rhythm
Answer
idioventricular rhythm
An idioventricular rhythm is an escape rhythm: the ventricles take over because the pacemakers above them have failed.
13.Evidence-based practice statistics
Answer: OR = 2.52
Likelihood ratios and risk measures: LR+ = sensitivity/(1 − specificity), LR− = (1 − sensitivity)/specificity. In a trial, EER and CER are the event rates with and without treatment: RR = EER/CER, ARR = CER − EER, RRR = ARR/CER and NNT = 1/ARR, rounded up. In a case-control study only the odds ratio OR = ad/(bc) can be used. An incidence rate = new cases/person-time.
Odds of exposure
cases: 157/53 = 2.962
controls: 227/193 = 1.176
Odds ratio
OR = (157·193)/(227·53) = 2.519 ≈ 2.52
Answer
OR = 2.52
OR above 1: exposure is more common among the cases.
14.Evidence-based practice statistics
Answer: p′ = 2.1 %
Predictive values depend on prevalence: PPV = Se·p/(Se·p + (1 − Sp)(1 − p)) and NPV = Sp(1 − p)/(Sp(1 − p) + (1 − Se)p). With likelihood ratios: pre-test odds = p/(1 − p), post-test odds = pre-test odds × LR, probability = odds/(1 + odds). For a population of N, true positives = N·p·Se and false positives = N·(1 − p)·(1 − Sp). NNT = 1/(CER × RRR), rounded up.